Problem 110 - Chemistry: The Central Science

Question

A sample of \(1.50 \mathrm{~g}\) of lead(II) nitrate is mixed with \(125 \mathrm{~mL}\) of \(0.100 \mathrm{M}\) sodium sulfate solution. (a) Write the chemical equation for the reaction that occurs. (b) Which is the limiting reactant in the reaction? (c) What are the concentrations of all ions that remain in solution after the reaction is complete?

Step-by-Step Solution

Answer

\[Pb(NO_3)_2(aq) + Na_2SO_4(aq) \rightarrow PbSO_4(s) + 2NaNO_3(aq)\] Lead(II) nitrate is the limiting reactant. After the reaction is complete, the remaining concentrations are: - [Na^+] = 0.200 M - [SO_4^2-] = 0.0638 M - [Pb^2+] = 0 M (precipitated as PbSO_4) - [NO_3^-] = 0.00906 mol (unchanged)

Step 1: Write the balanced chemical equation

Combine lead(II) nitrate, Pb(NO3)2, with sodium sulfate, Na2SO4, to form lead(II) sulfate, PbSO4, and sodium nitrate, NaNO3. Write the balanced chemical equation for this reaction: \[Pb(NO_3)_2(aq) + Na_2SO_4(aq) \rightarrow PbSO_4(s) + 2NaNO_3(aq)\]

Step 2: Determine the limiting reactant

To find the limiting reactant, we need to calculate the moles of lead(II) nitrate and sodium sulfate. Given: Mass of lead(II) nitrate: 1.50 g Molar mass of lead(II) nitrate: 331 g/mol mol of lead(II) nitrate = mass/molar mass mol of lead(II) nitrate = \( \frac{1.50 g}{331 g/mol} \) mol of lead(II) nitrate = 0.00453 mol Volume of sodium sulfate solution: 125 mL = 0.125 L Molarity of sodium sulfate solution: 0.100 M mol of sodium sulfate = molarity × volume mol of sodium sulfate = 0.100 mol/L × 0.125 L mol of sodium sulfate = 0.0125 mol Now, we must find the stoichiometric ratio of the two reactants: \( \frac{mol Pb(NO_3)_2}{mol Na_2SO_4} = \frac{0.00453}{0.0125} = 0.362 \) Since the stoichiometric ratio of the balanced equation is 1:1, lead(II) nitrate is the limiting reactant.

Step 3: Calculate the concentrations of remaining ions

From step 2, we know that lead(II) nitrate is the limiting reactant and that 0.00453 mol of lead(II) nitrate react. In the balanced equation, one mole of lead(II) nitrate reacted with one mole of sodium sulfate. So, 0.00453 mol of sodium sulfate also reacted, leaving: 0.0125 mol - 0.00453 mol = 0.00797 mol sodium sulfate remaining Since no sodium ions are consumed in the reaction, the initial concentration of sodium ions remains the same. The initial concentration of sodium ions can be calculated as: \[ [Na^+] = 2 \times M(Na_2SO_4) = 2 \times 0.100M = 0.200M \] For sulfate ions, we calculate the remaining concentration after the reaction: \[ [SO_4^{2-}] = \frac{0.00797 mol}{0.125 L} = 0.0638 \mathrm{~M} \] For lead(II) nitrate, since it is the limiting reactant, all of it reacts, leaving no lead ions in solution. For nitrate ions, the concentration remains the same, as it's not involved in the reaction: \[ [NO_3^-] = 2 \times M(Pb (NO_3)_2) = 2 \times 0.00453 mol \] \[ [NO_3^-] = 0.00906 mol \] After the reaction is complete, the following remaining concentrations are found: - [Na^+] = 0.200 M - [SO_4^2-] = 0.0638 M - [Pb^2+] = 0 M (precipitated as PbSO_4) - [NO_3^-] = 0.00906 mol (unchanged)